POJ 1742 Coins(多重背包可行性问题)

People in Silverland use coins.They have coins of value A1,A2,A3…An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn’t know the exact price of the watch.

You are to write a program which reads n,m,A1,A2,A3…An and C1,C2,C3…Cn corresponding to the number of Tony’s coins of value A1,A2,A3…An then calculate how many prices(form 1 to m) Tony can pay use these coins.

Input

The input contains several test cases. The first line of each test case contains two integers n(1<=n<=100),m(m<=100000).The second line contains 2n integers, denoting A1,A2,A3…An,C1,C2,C3…Cn (1<=Ai<=100000,1<=Ci<=1000). The last test case is followed by two zeros.

Output

For each test case output the answer on a single line.

Sample Input

3 10

1 2 4 2 1 1

2 5

1 4 2 1

0 0

Sample Output

8

4

题意：你有N种硬币，每种硬币有价值和个数，问你用这N种硬币，能够凑出1-M这些钱的多少个。

思路：一种类似于多重背包的题目，不过是求凑成价值的个数，和求多少种取法的问题不同。dp[i][j]表示用前i种硬币能不能呢凑出j这种价值。num[i][j];表示用前i种硬币凑成价值j,时第i种用了多少枚。

代码如下：

#include<iostream>
#include<string>
#include<cmath>
#include<vector>
#include<algorithm>
#include<set>
#include<queue>

using namespace std;
const int MAX = 100010;
const int MAX_N = 110;
bool have[MAX];
int num[MAX];
pair<int,int> p[MAX_N];
int N,M;
void solve(){
memset(have,false,sizeof(have));
have[0] = true;
for(int i=1;i<=N;++i){
memset(num,0,sizeof(num));//当前硬币没有没使用所以全部置为0，
for(int j=p[i].first;j<=M;++j){
if(!have[j] && have[j-p[i].first] && num[j-p[i].first] < p[i].second){
//表示这一种方案可以凑出来。
have[j] = true;
num[j] = num[j-p[i].first] + 1;//表示当前这种方案又是使用了一枚第i个硬币。
}
}
}
int res = 0;
for(int i=1;i<=M;++i)
if(have[i]){
res++;
}
cout << res << endl;
}
int main(void){
while(cin >> N >> M){
if(N == 0 && M == 0)    break;
for(int i=1;i<=N;++i)
cin >> p[i].first;
for(int i=1;i<=N;++i)
cin >> p[i].second;
solve();
}

return 0;
}